Mix Examples- States of Matter Questions in English

(C) The total pressure of a moist gas is the sum of the partial pressure of the dry gas and the vapour pressure of water: $P_{total} = P_{gas} + P_{H_2O}$.
Given $P_{total} = 4 \ atm$ and $P_{H_2O} = 0.4 \ atm$,so $P_{gas} = 4 - 0.4 = 3.6 \ atm$.
According to Boyle's Law,at constant temperature,$P_1V_1 = P_2V_2$. When the volume is doubled $(V_2 = 2V_1)$,the new partial pressure of the dry gas becomes $P_{gas, new} = P_{gas} / 2 = 3.6 / 2 = 1.8 \ atm$.
The vapour pressure of water remains constant at a given temperature,so $P_{H_2O} = 0.4 \ atm$.
The new total pressure is $P_{total, new} = 1.8 + 0.4 = 2.2 \ atm$.
Since $2.2 \ atm = 22 \times 10^{-1} \ atm$,the value of $x$ is $22$.

(A) Given: $P = 6 \ bar = 6 \times 10^{5} \ Pa$,$V = 0.0125 \ m^{3}$,$T = 27 + 273 = 300 \ K$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$.
Using the ideal gas equation $PV = n_{mix}RT$:
$n_{mix} = \frac{PV}{RT} = \frac{6 \times 10^{5} \times 0.0125}{8.3 \times 300} = \frac{7500}{2490} \approx 3.012 \ mol \approx 3 \ mol$.
Let the moles of $He$ be $x$ and moles of $H_{2}$ be $(3-x)$.
The total mass is $10 \ g$,so: $4x + 2(3-x) = 10$.
$4x + 6 - 2x = 10 \implies 2x = 4 \implies x = 2 \ mol$.
Mass of $He = n \times M = 2 \ mol \times 4 \ g \ mol^{-1} = 8 \ g$.

(D) The initial pressure of helium in both balloons $A$ and $B$ is given as $2.0 \ atm$.
Since the temperature is constant and the pressure in both balloons is identical,connecting them will not result in any net flow of gas between the balloons.
Therefore,the final pressure of $He$ in the system remains $2.0 \ atm$.

(D) Statement $A$ is incorrect because surface tension arises from unbalanced attractive forces on the surface,not equal forces in the bulk.
Statement $B$ is correct because molecules on the surface experience a net inward pull,leading to surface tension.
Statement $C$ is incorrect because molecules in the bulk are in constant motion and can reach the surface due to their kinetic energy.
Statement $D$ is correct because molecules on the surface have higher energy and can escape into the gas phase,creating vapour pressure in a closed system.
Therefore,there are $2$ correct statements ($B$ and $D$).

(D) . At high pressure $(200 \ atm)$,$H_2$ behaves as a real gas where $Z \neq 1$ and the volume correction term $V_m-b$ becomes significant,leading to $P(V_m-b)=RT$.
$B$. At $P \approx 0$,all real gases approach ideal behavior,so $PV=nRT$.
$C$. $CO_2$ is a polarizable molecule with significant intermolecular attractive forces,so $Z \neq 1$ and attractive forces are dominant.
$D$. At very large molar volume,real gases approach ideal behavior,so $PV=nRT$.

(B) $1.$ In the $P-V$ diagram:
$K \rightarrow L$: Constant pressure $(P)$,volume $(V)$ increases. By Charles's Law $(V \propto T)$,$T$ increases (Heating).
$L \rightarrow M$: Constant volume $(V)$,pressure $(P)$ decreases. By Gay-Lussac's Law $(P \propto T)$,$T$ decreases (Cooling).
$M \rightarrow N$: Constant pressure $(P)$,volume $(V)$ decreases. By Charles's Law,$T$ decreases (Cooling).
$N \rightarrow K$: Constant volume $(V)$,pressure $(P)$ increases. By Gay-Lussac's Law,$T$ increases (Heating).
Thus,the sequence is Heating,Cooling,Cooling,Heating,which corresponds to option $(C)$.
$2.$ An isochoric process is one where volume remains constant. In the diagram,the vertical lines represent constant volume processes.
These are $L \rightarrow M$ and $N \rightarrow K$. This corresponds to option $(B)$.
Therefore,the correct answer is $(C, B)$.

(D) The phase diagram of $H_2O$ shows that at $273.15 \ K$ and $1 \ atm$,ice and water are in equilibrium.
Since the melting point of ice decreases with an increase in pressure (due to the negative slope of the solid-liquid equilibrium line for $H_2O$),increasing the pressure at a constant temperature of $273.15 \ K$ shifts the system into the liquid region.
Therefore,the solid phase (ice) will melt and disappear completely,converting into the liquid phase (water).

(C) Let the mass of each gas be $x \ g$.
$n_{H_2} = \frac{x \ g}{2 \ g \ mol^{-1}} = \frac{x}{2} \ mol$.
$n_{He} = \frac{x \ g}{4 \ g \ mol^{-1}} = \frac{x}{4} \ mol$.
According to Dalton's law of partial pressures,the partial pressure of a gas is directly proportional to its number of moles in a mixture at constant temperature and volume $(P_i = \frac{n_i RT}{V})$.
Therefore,the ratio of partial pressures is equal to the ratio of the number of moles:
$\frac{P_{H_2}}{P_{He}} = \frac{n_{H_2}}{n_{He}} = \frac{x/2}{x/4} = \frac{4}{2} = \frac{2}{1}$.
Thus,the ratio is $2: 1$.

(A) The ideal gas equation is $pV = nRT$.
Since $V$ and $T$ are constant,$p \propto n$.
Therefore,the substance with the least number of moles $(n)$ will exert the least pressure.
Note: The molar mass of chlorine gas $(Cl_2)$ is $71 \ g/mol$.
$(A)$ $n = \frac{0.0355 \ g}{71 \ g/mol} = 0.0005 \ mol = 5 \times 10^{-4} \ mol$.
$(B)$ $n = \frac{0.071 \ g}{71 \ g/mol} = 0.001 \ mol = 1 \times 10^{-3} \ mol$.
$(C)$ $n = \frac{6.023 \times 10^{21}}{6.023 \times 10^{23}} = 0.01 \ mol$.
$(D)$ $n = 0.02 \ mol$.
Comparing the values,$5 \times 10^{-4} \ mol$ is the smallest.
Thus,$0.0355 \ g$ of chlorine exerts the least pressure.

(D) Given,$T_{1} = 273 + 10 = 283 \ K$ and $T_{2} = 273 + 20 = 293 \ K$.
Average kinetic energy $(KE)$ is directly proportional to temperature $(T)$: $KE = \frac{3}{2} RT$.
Ratio $\frac{KE_{2}}{KE_{1}} = \frac{293}{283} \approx 1.035$.
Root mean square velocity $(v_{rms})$ is proportional to $\sqrt{T}$: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Ratio $\frac{v_{rms,2}}{v_{rms,1}} = \sqrt{\frac{293}{283}} \approx 1.017$.
Since the temperature change is small relative to the absolute temperature,both the average kinetic energy and the rms velocity increase,but not significantly.

(D) The graph represents the cooling curve of a mixture of two liquids $A$ and $B$.
At point $B$ $(-25^{\circ} C)$,liquid $B$ starts to solidify. Along the curve $BC$,liquid $B$ is solidifying,so two phases (liquid $B$ and solid $B$) co-exist along with liquid $A$.
At point $C$ $(-78^{\circ} C)$,liquid $A$ also starts to solidify.
Along the line $CD$,both liquid $A$ and liquid $B$ are undergoing solidification.
Therefore,the phases present are: liquid $A$,solid $A$,liquid $B$,and solid $B$.
Thus,$4$ different phases co-exist along the line $CD$.

(A) The boiling point is inversely proportional to the vapour pressure at a given temperature.
Given boiling points: $C (308 \ K) < A (350 \ K) < B (373 \ K)$.
Therefore,the order of vapour pressure at any temperature is $C > A > B$.
$I.$ Since $B$ has the highest boiling point,it has the lowest vapour pressure,meaning it remains primarily in the liquid phase in vessel $(I)$. Thus,vessel $(I)$ is rich in liquid $B$. This statement is correct.
$II.$ Since $C$ has the lowest boiling point,it has the highest vapour pressure,meaning it vaporizes most easily. Thus,vessel $(II)$ will be rich in the vapour of $C$. This statement is correct.
$III.$ As established,the order of vapour pressure is $C > A > B$. This statement is correct.
All statements $I, II,$ and $III$ are correct.

(C) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the vapour density $(VD)$:
$r \propto \frac{1}{\sqrt{VD}}$
Therefore,$\frac{r_A}{r_B} = \sqrt{\frac{VD_B}{VD_A}}$
Given that $r_A = \frac{20 \ mL}{1 \ min} = 20 \ mL/min$ and $r_B = \frac{10 \ mL}{1 \ min} = 10 \ mL/min$.
Substituting the values:
$\frac{20}{10} = \sqrt{\frac{x}{VD_A}}$
$2 = \sqrt{\frac{x}{VD_A}}$
Squaring both sides:
$4 = \frac{x}{VD_A}$
$VD_A = \frac{x}{4}$

(C) Statement-$I$ is correct: Thermal energy promotes random motion of particles,while intermolecular forces hold them together. If thermal energy dominates,the particles move apart,leading to a gaseous state.
Statement-$II$ is correct: From the ideal gas equation,$PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$. Rearranging gives $P = \frac{m}{V} \times \frac{RT}{M}$,where $\frac{m}{V} = d$ (density). Thus,$P = d \times \frac{RT}{M}$. At constant temperature $(T)$ and molar mass $(M)$,$P \propto d$ or $d \propto P$.

(D) Initial moles of $He \ (n_1) = 1 \ mol$.
Initial moles of $CH_4 \ (n_2) = 1 \ mol$.
Moles of $He$ escaped $(n_1') = \frac{2 \ L}{22.4 \ L/mol} \approx 0.089 \ mol$.
Moles of $CH_4$ escaped $(n_2') = \frac{1 \ L}{22.4 \ L/mol} \approx 0.044 \ mol$.
Remaining moles of $He \ (n_1'') = 1 - 0.089 = 0.911 \ mol$.
Remaining moles of $CH_4 \ (n_2'') = 1 - 0.044 = 0.956 \ mol$.
Total remaining moles $= 0.911 + 0.956 = 1.867 \ mol$.
Mole fraction of $He \ (\chi_{He}) = \frac{0.911}{1.867} \approx 0.488$.
Mole fraction of $CH_4 \ (\chi_{CH_4}) = \frac{0.956}{1.867} \approx 0.512$.

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$.
Therefore,$\frac{r_1}{r_2} = \frac{V_1/t_1}{V_2/t_2} = \sqrt{\frac{M_2}{M_1}}$.
Given: $V_1 = 60 \ cm^3, M_1 = 64 \ g \ mol^{-1}, t_1 = x$.
$V_2 = 360 \ cm^3, M_2 = 4 \ g \ mol^{-1}, t_2 = y$.
Substituting the values: $\frac{60/x}{360/y} = \sqrt{\frac{4}{64}}$.
$\frac{60}{x} \times \frac{y}{360} = \frac{2}{8} = \frac{1}{4}$.
$\frac{y}{6x} = \frac{1}{4}$.
$\frac{y}{x} = \frac{6}{4} = \frac{3}{2}$.
Therefore,the ratio $\frac{x}{y} = \frac{2}{3}$.

(C) The kinetic energy $(KE)$ of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$.
Since the temperature $(T)$ is constant,the kinetic energy is directly proportional to the number of moles $(n)$,i.e.,$KE \propto n$.
Therefore,the ratio of kinetic energies is: $\frac{KE_{H_2}}{KE_{O_2}} = \frac{n_{H_2}}{n_{O_2}}$.
Calculate the number of moles for each gas:
$n_{H_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{3 \ g}{2 \ g/mol} = 1.5 \ mol$.
$n_{O_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \ g}{32 \ g/mol} = 0.125 \ mol$.
Now,find the ratio: $\frac{KE_{H_2}}{KE_{O_2}} = \frac{1.5}{0.125} = \frac{12}{1}$.
Thus,the ratio is $12: 1$.

(D) For an ideal gas,the equation of state is $PV = nRT$. For $1 \ mol$ of gas,$PV = RT$,therefore the compressibility factor $Z = \frac{PV}{RT} = 1$. Thus,statement $(A)$ is correct.
Uranium isotopes (${ }^{235}U$ and ${ }^{238}U$) are separated by converting them into $UF_6$ vapours,which are then separated based on their different rates of diffusion (Graham's Law). Thus,statement $(B)$ is correct.
Kinetic energy $(KE)$ of gas molecules is directly proportional to the absolute temperature $(T)$,given by the relation $KE = \frac{3}{2}RT$. Therefore,a decrease in temperature leads to a decrease in the kinetic energy of gas molecules. Thus,statement $(C)$ is incorrect.
Therefore,statements $(A)$ and $(B)$ are correct.

(B) The number of moles of hydrogen is calculated as $n_{H_2} = \frac{3 \ g}{2 \ g \ mol^{-1}} = 1.5 \ mol$.
The number of moles of oxygen is calculated as $n_{O_2} = \frac{80 \ g}{32 \ g \ mol^{-1}} = 2.5 \ mol$.
The total number of moles in the gaseous mixture is $n = 1.5 + 2.5 = 4 \ mol$.
The total kinetic energy of an ideal gas mixture is given by the formula $KE = \frac{3}{2} n R T$.
Substituting the values,we get $KE = \frac{3}{2} \times 4 \times R T = 6RT$.

(B) $Statement-I$ is incorrect. If intermolecular forces are stronger than thermal energy,the substance prefers to be in a solid or liquid state,not a gaseous state. Thermal energy promotes random motion,which leads to the gaseous state.
$Statement-II$ is correct. There are $11$ elements that are gases at room temperature $(H_2, He, N_2, O_2, F_2, Ne, Cl_2, Ar, Kr, Xe, Rn)$. Wait,checking the count: $H, He, N, O, F, Ne, Cl, Ar, Kr, Xe, Rn$ equals $11$ elements. Therefore,the statement claiming $10$ is incorrect.
Both statements are incorrect.

(B) The critical temperature of $CO_2$ is $31.1^{\circ} C$ $(304.1 \ K)$,not $27.5^{\circ} C$. Therefore,statement $B$ is incorrect.
At Boyle temperature,the effects of attractive and repulsive forces balance out,making the gas behave ideally over a range of pressure.
Above the critical temperature,gases cannot be liquefied by pressure alone,and they exhibit behavior closer to ideal gases as temperature increases.
For $H_2$ and $He$,the compressibility factor $Z$ is always greater than $1$ at room temperature because the repulsive forces dominate due to their small molecular size.

(C) $I$. Compressibility factor $(Z)$ is defined as the ratio of $PV$ to $nRT$. For an ideal gas,$PV = nRT$,therefore $Z = 1$ at all temperatures and pressures. Thus,statement $I$ is correct.
$II$. The average kinetic energy of one mole of an ideal gas is given by $K.E. = \frac{3}{2}RT$. Since it depends only on the temperature $(T)$ and the gas constant $(R)$,it is independent of the molar mass of the gas. Therefore,the kinetic energy of $NO_{(g)}$ and $N_2O_{4(g)}$ at the same temperature $T$ must be equal. Thus,statement $II$ is incorrect.
$III$. According to Graham's law of diffusion,the rate of diffusion $(r)$ of a gas is inversely proportional to the square root of its density $(d)$,i.e.,$r \propto \frac{1}{\sqrt{d}}$. Thus,statement $III$ is correct.
Conclusion: Statements $I$ and $III$ are correct.

(A) Statement-$I$ is correct: The compressibility factor $Z$ is defined as $Z = \frac{V_{m, \text{real}}}{V_{m, \text{ideal}}}$,which represents the ratio of the molar volume of a real gas to that of an ideal gas at the same temperature and pressure.
Statement-$II$ is correct: The root mean square $(RMS)$ velocity of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$. Since $R$ and $M$ are constants for a given gas,$v_{rms} \propto \sqrt{T}$.
Therefore,both statements are correct.

(B) $I$. Glass is considered a supercooled liquid or an extremely viscous liquid,which is a correct statement.
$II$. Surface tension of liquids decreases with an increase in temperature because the kinetic energy of molecules increases,which weakens the intermolecular forces of attraction. This is a correct statement.
$III$. The compressibility factor $(Z)$ for an ideal gas is defined as $Z = \frac{PV}{nRT} = 1$. Therefore,the statement that it is zero is incorrect.

(A) Viscosity of a liquid is a measure of its resistance to flow.
In liquids,viscosity decreases as temperature increases because the kinetic energy of the molecules increases,allowing them to overcome the intermolecular attractive forces.
Therefore,Statement-$I$ is correct.
The $SI$ unit of the coefficient of viscosity $(\eta)$ is $N \ s \ m^{-2}$ or $Pa \ s$ (Pascal-second).
Therefore,Statement-$II$ is also correct.

(D) The correct answer is $D$.
Surface tension causes liquid droplets to take a spherical shape to minimize their surface area for a given volume.
However,on a flat surface,the gravitational force and the adhesive forces between the liquid and the surface cause the droplet to flatten,making it non-spherical.
Therefore,the statement that liquid droplets are perfectly spherical on a flat surface is incorrect.

(B) Evaporation is a surface phenomenon.
When the surface area decreases,fewer molecules are exposed to the surface,leading to a decrease in the rate of evaporation.
Therefore,the statement that evaporation increases with a decrease in surface area is incorrect.

(D) The thrust force $(T_h)$ exerted by a liquid on a submerged object is given by the pressure $(P)$ multiplied by the area $(A)$.
$T_h = P \times A$
Since both discs are at the same depth $(h)$,the pressure $P = h \rho g$ is the same for both.
Therefore,$T_h \propto A$.
$T_A : T_B = A_A : A_B = \pi (r_A)^2 : \pi (r_B)^2$
$T_A : T_B = (2)^2 : (4)^2 = 4 : 16 = 1 : 4$.

(B) First,calculate the number of moles of each gas:
$n_{CH_4} = \frac{3.2 \ g}{16 \ g/mol} = 0.2 \ mol$
$n_{CO_2} = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$
Total moles $n_{total} = 0.2 + 0.1 = 0.3 \ mol$
Using the ideal gas equation $PV = nRT$:
$P = \frac{n_{total}RT}{V}$
Given $T = 27^{\circ}C = 300 \ K$,$V = 9 \ dm^3 = 9 \ L$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$:
$P = \frac{0.3 \times 0.0821 \times 300}{9} \ atm$
$P = \frac{7.389}{9} \ atm = 0.821 \ atm$
Thus,the total pressure is approximately $0.82 \ atm$.

(C) In laminar flow over a fixed surface,the layer of liquid in direct contact with the surface has the lowest velocity due to friction and adhesive forces.
As the distance from the fixed surface increases,the velocity of the liquid layers increases.
Layer $1$ is closest to the surface,followed by Layer $2$,and then Layer $3$.
Therefore,the velocity order is $V_3 > V_2 > V_1$.

(D) Statements $1, 2$ and $3$ are correct.
Statement $4$ is incorrect because the magnitude of surface tension is directly proportional to the strength of intermolecular attractive forces. Stronger attractive forces between molecules lead to a higher magnitude of surface tension.

(B) The total kinetic energy $(K.E.)$ of an ideal gas is given by the formula: $K.E. = \frac{3}{2} nRT$.
For flask $A$ containing $H_2$: $n_A = \frac{m}{M_{H_2}} = \frac{m}{2}$ and $T_A = 300 \ K$.
Thus,$(K.E.)_A = \frac{3}{2} \times \frac{m}{2} \times R \times 300$.
For flask $B$ containing $CO_2$: $n_B = \frac{m}{M_{CO_2}} = \frac{m}{44}$ and $T_B = 600 \ K$.
Thus,$(K.E.)_B = \frac{3}{2} \times \frac{m}{44} \times R \times 600$.
Taking the ratio: $\frac{(K.E.)_A}{(K.E.)_B} = \frac{\frac{m}{2} \times 300}{\frac{m}{44} \times 600} = \frac{300/2}{600/44} = \frac{150}{600/44} = \frac{150 \times 44}{600} = \frac{44}{4} = 11:1$.

(C) Helium $(He)$ and Neon $(Ne)$ are both monoatomic gases.
For a monoatomic gas,the degree of freedom $(f)$ is $3$.
The ratio of molar specific heat capacities $\gamma = \frac{C_p}{C_V} = 1 + \frac{2}{f}$.
Since both gases are monoatomic,the mixture will also behave as a monoatomic gas with $f = 3$.
Therefore,$\frac{C_p}{C_V} = 1 + \frac{2}{3} = \frac{5}{3}$.

(B) The variation of the viscosity coefficient $(\eta)$ of liquids with temperature $(T)$ is described by the Andrade equation,which is given by the expression: $\eta = A e^{E / R T}$.
Here,$A$ is a constant,$E$ is the activation energy for viscous flow,$R$ is the universal gas constant,and $T$ is the absolute temperature.
As temperature increases,the viscosity of liquids decreases,which is consistent with the exponential form $\eta = A e^{E / R T}$.

(B) Using the ideal gas equation $PV = nRT$,the number of moles of oxygen in flask '$A$' is $n_{O_2} = \frac{P_A V_A}{RT} = \frac{5 \times V}{RT}$.
Similarly,the number of moles of helium in flask '$B$' is $n_{He} = \frac{P_B V_B}{RT} = \frac{3 \times 2}{RT} = \frac{6}{RT}$.
The mole fraction of oxygen in the mixture is given by $x_{O_2} = \frac{n_{O_2}}{n_{O_2} + n_{He}} = 0.2$.
Substituting the values: $\frac{5V/RT}{5V/RT + 6/RT} = 0.2$.
This simplifies to $\frac{5V}{5V + 6} = 0.2$.
$5V = 0.2(5V + 6) \implies 5V = V + 1.2$.
$4V = 1.2 \implies V = 0.3 \ L$.

(B) Using the ideal gas equation,$PV = nRT$,for $1 \ mol$ of the mixture,the volume $V$ is given by $V = \frac{RT}{P}$.
Given $P = 760 \ torr = 1 \ atm$,$T = 300 \ K$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,we have $V = \frac{0.0821 \times 300}{1} = 24.63 \ L$.
The total mass of $24.63 \ L$ of the mixture is $M = V \times d = 24.63 \ L \times 0.543 \ g \ L^{-1} = 13.37 \ g$.
Let $x$ be the mole fraction of $O_2$. Then the mole fraction of $He$ is $(1 - x)$.
The average molar mass of the mixture is $M_{avg} = x(32) + (1 - x)(4) = 13.37$.
Solving for $x$: $32x + 4 - 4x = 13.37$ $\Rightarrow 28x = 9.37$ $\Rightarrow x \approx 0.3346$.
The mass of $O_2$ in $1 \ mol$ of the mixture is $0.3346 \times 32 \approx 10.707 \ g$.
The mass percent of $O_2$ is $\frac{\text{mass of } O_2}{\text{total mass}} \times 100 = \frac{10.707}{13.37} \times 100 \approx 80.08 \% \approx 80 \%$.

(B) The correct matches are as follows:
$(A)$ Viscosity: The $SI$ unit of viscosity is $kg \ m^{-1} \ s^{-1}$ $(V)$.
$(B)$ Ideal gas behaviour: It is described by the compressibility factor $(Z = PV/nRT)$,which is $1$ for ideal gases $(III)$.
$(C)$ Liquefaction of gases: This is related to the critical temperature $(T_c)$ of a gas $(I)$.
$(D)$ Charles' law: It describes the relationship between volume and temperature at constant pressure,represented by isobars on a graph $(II)$.
Therefore,the correct matching is $A-V, B-III, C-I, D-II$.

(A) According to Graham's law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the density $(\rho)$ of the gas: $r \propto \frac{1}{\sqrt{\rho}}$.
From the ideal gas equation,$PV = nRT = \frac{m}{MW} RT$,where $m$ is the mass and $MW$ is the molecular weight.
Rearranging gives $\frac{m}{V} = \rho = \frac{P \times MW}{RT}$.
Since $T$ is constant,$\rho \propto P \times MW$.
Therefore,the rate of diffusion is given by $r \propto \frac{1}{\sqrt{P \times MW}}$.
For two gases,$\frac{r_{CH_4}}{r_X} = \sqrt{\frac{P_X \times (MW)_X}{P_{CH_4} \times (MW)_{CH_4}}}$.
Given $r_{CH_4} = 2r_X$,$P_{CH_4} = 1.0 \ atm$,$P_X = 1.45 \ atm$,and $(MW)_{CH_4} = 16 \ g/mol$:
$2 = \sqrt{\frac{1.45 \times (MW)_X}{1.0 \times 16}}$.
Squaring both sides: $4 = \frac{1.45 \times (MW)_X}{16}$.
$(MW)_X = \frac{4 \times 16}{1.45} = \frac{64}{1.45} \approx 44.13 \approx 44$.

(B) From Graham's law of diffusion,$\frac{r_{CH_4}}{r_X} = \sqrt{\frac{M_X}{M_{CH_4}}}$.
Given that $r_{CH_4} = 2 \cdot r_X$,we have $2 = \sqrt{\frac{M_X}{16}}$.
Squaring both sides,$4 = \frac{M_X}{16}$,so $M_X = 64 \ g/mol$.
The number of moles of gas $X$ in $32 \ g$ is $n = \frac{32}{64} = 0.5 \ mol$.
The number of molecules is $n \times N = 0.5 \times N = \frac{N}{2}$.

(A) Let the weight of both gases $A$ and $B$ be $w \text{ g}$.
Given molecular weight ratio $M_A : M_B = 1 : 4$. Let $M_A = M$ and $M_B = 4M$.
Moles of $A$ $(n_A)$ $= \frac{w}{M}$.
Moles of $B$ $(n_B)$ $= \frac{w}{4M}$.
Mole ratio $n_A : n_B = \frac{w}{M} : \frac{w}{4M} = 4 : 1$.
Partial pressure of $B$ $(p_B)$ $= \text{Mole fraction of } B \times P_{\text{total}}$.
$p_B = \frac{n_B}{n_A + n_B} \times P = \frac{1}{4 + 1} \times P = \frac{P}{5} \text{ atm}$.

(D) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ and directly proportional to the volume $V$ diffused per unit time $t$: $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$.
Given for Helium $(He)$: $V_1 = 500 \ mL$,$t_1 = 30 \ min$,$M_1 = 4 \ g/mol$.
Given for Sulfur dioxide $(SO_2)$: $V_2 = 1000 \ mL$,$t_2 = t$,$M_2 = 64 \ g/mol$.
The ratio is given by: $\frac{r_1}{r_2} = \frac{V_1/t_1}{V_2/t_2} = \sqrt{\frac{M_2}{M_1}}$.
Substituting the values: $\frac{500/30}{1000/t} = \sqrt{\frac{64}{4}}$.
$\frac{500}{30} \times \frac{t}{1000} = \sqrt{16} = 4$.
$\frac{t}{60} = 4$.
$t = 240 \ minutes$.
Converting to hours: $t = \frac{240}{60} = 4 \ hours$.

(D) Temperature of helium $= 27^{\circ} C = 300 \ K$.
The root-mean-square velocity of helium is given by $\mu_{rms} = \sqrt{\frac{3RT}{M_{He}}} = \sqrt{\frac{3R \times 300}{4}}$.
The most probable speed of the unknown gas is $\mu_{mp} = \sqrt{\frac{2RT}{M}} = \sqrt{\frac{2R \times 7200}{M}}$.
Equating $\mu_{rms} = \mu_{mp}$:
$\sqrt{\frac{3R \times 300}{4}} = \sqrt{\frac{2R \times 7200}{M}}$.
Squaring both sides: $\frac{900R}{4} = \frac{14400R}{M}$.
$225 = \frac{14400}{M}$.
$M = \frac{14400}{225} = 64 \ g/mol$.
The molar mass of $SO_2$ is $32 + 2 \times 16 = 64 \ g/mol$.
Therefore,the gas is $SO_2$.

(B) The kinetic energy for $n$ moles of an ideal gas is given by $K.E. = \frac{3}{2} pV$.
Given: $K.E. = 98.03 \ L \ atm$ and $V = 24.6 \ L$.
Substituting the values: $98.03 = \frac{3}{2} \times p \times 24.6$.
$p = \frac{98.03 \times 2}{3 \times 24.6} = \frac{196.06}{73.8} \approx 2.66 \ atm$.
Since the reaction $A_{(g)} + B_{(g)} \longrightarrow 2 D_{(g)}$ involves $2 \text{ moles}$ of reactants producing $2 \text{ moles}$ of products,the total number of moles remains constant ($2 \text{ moles}$ total).
Thus,the pressure exerted by the gas $D$ at the end of the reaction is $2.66 \ atm$.

(C) The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = nRT$, where $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature in Kelvin.
Given:
For helium: $n_{He} = 0.3 \text{ mol}$
For argon: $n_{Ar} = 0.4 \text{ mol}$, $T_{Ar} = 400 \text{ K}$
According to the problem, $KE_{He} = KE_{Ar}$.
Substituting the values:
$0.3 \times R \times T = 0.4 \times R \times 400$
Dividing both sides by $R$:
$0.3 \times T = 160$
$T = \frac{160}{0.3} = 533.33 \text{ K} \approx 533 \text{ K}$.

(A) Given: $P = 1 \ atm$,$T = 400 \ K$,$\delta = 0.920 \ g \ L^{-1}$,$R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$.
Using the ideal gas equation $PV = nRT$ and $n = \frac{m}{M}$,we get $PM = \delta RT$.
$M = \frac{\delta RT}{P} = \frac{0.920 \times 0.082 \times 400}{1} = 30.176 \ g \ mol^{-1}$.
Let $x$ be the mole fraction of $N_2$. Then the mole fraction of $O_2$ is $(1-x)$.
The average molar mass $M = x M_{N_2} + (1-x) M_{O_2}$.
$30.176 = x(28) + (1-x)(32)$.
$30.176 = 28x + 32 - 32x$.
$4x = 32 - 30.176 = 1.824$.
$x = \frac{1.824}{4} = 0.456$.
Thus,the mole fraction of nitrogen is $0.456$.